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fredd [130]
3 years ago
7

Ron invested $700 for 9 years at 3.4% compounded semi-annually. How many compounding periods are there over the term of the inve

stment?
Mathematics
1 answer:
Neko [114]3 years ago
3 0

Answer:

financial equations use the formula

A=P (1 + \frac{r}n)} ^{nt}

where

A is accrued amount

P = principal

r= interest rate

t = number of periods (NOTE: NOT NUMBER OF YEARS !!!)

r = interest rate (NOTE: APR the whole years interest NOT the period)

So in your question the compounding is being done "Semi-annually:

NOTE:

Annual = 1

semi-annual = 2

Quarterly = 4

monthly = 12

So your number of periods is ..... number of years (9) times the number of periods in a year (2 - for semi-annual)

9 * 2 = 18 there are 18 periods

A=P (1 + \frac{.034}2)} ^{18}

Step-by-step explanation:

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Kobotan [32]

Answer:

The answer is A

Step-by-step explanation:

5 0
3 years ago
X - 4 = 9<br> Can someone please help me
mihalych1998 [28]

Answer:

x=13

Step-by-step explanation:

isolate the x by adding4 to each side of the equation

x - 4 = 9

 + 4 =+ 4

and then the -4 cancels out and you get x = 13

I hope this helps :)

7 0
3 years ago
Read 2 more answers
Y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′′(0) = 0
Snowcat [4.5K]

Answer:

y(t) =  3u₂(t) [ e^{-2t+4}  - e^{-5t + 10)} ] - 4u₅(t) [ e^{-2t+10)}  - e^{-5t + 25)} ]

Step-by-step explanation:

To find - y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′(0) = 0

Formula used -

L{δ(t − c)} = e^{-cs}

L{f''(t) = s²F(s) - sf(0) - f'(0)

L{f'(t) = sF(s) - f(0)

Solution -

By Applying Laplace transform, we get

L{y″+ 5y′ + 6y} = L{3δ(t − 2) − 4δ(t −5)}

⇒L{y''} + 5L{y'} + 6L{y} = 3L{δ(t − 2)}  − 4L{δ(t −5)}

⇒s²Y(s) - sy(0) - y'(0) + 5[sY(s) - y(0)] + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒s²Y(s) - 0 - 0 + 5[sY(s) - 0] + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒s²Y(s) + 5sY(s) + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s² + 5s + 6] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s² + 3s + 2s + 6] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s(s + 3) + 2(s + 3)] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[(s + 2)(s + 3)] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒Y(s) = \frac{3e^{-2s} }{(s + 2)(s + 3)} -  \frac{4e^{-5s} }{(s + 2)(s + 3)}

Now,

Let

\frac{1}{(s+2)(s+3)} = \frac{A}{s+2}  + \frac{B}{s+3} \\\frac{1}{(s+2)(s+3)} = \frac{A(s + 3) + B(s+2)}{(s+2)(s+3)}\\1 = As + 3A + Bs + 2B\\1 = (A+B)s + (3A + 2B)

By Comparing, we get

A + B = 0 and 3A + 2B = 1

⇒A = -B

and

3(-B) + 2B = 1

⇒-B = 1

⇒B = -1

So,

A = 1

∴ we get

\frac{1}{(s+2)(s+3)} = \frac{1}{s+2}  + \frac{-1}{s+3}

So,

Y(s) = 3e^{-2s}[ \frac{1}{(s + 2)} -    \frac{1}{(s + 3)}] - 4e^{-5s}[ \frac{1}{(s + 2)} -    \frac{1}{(s + 3)}]

⇒Y(s) = 3e^{-2s} \frac{1}{(s + 2)} -    3e^{-2s} \frac{1}{(s + 3)} - 4e^{-5s}\frac{1}{(s + 2)} + 4e^{-5s}\frac{1}{(s + 3)}

By applying inverse Laplace , we get

y(t) = 3u₂(t) [ e^{-2(t-2)}  - e^{-5(t - 2)} ] - 4u₅(t) [ e^{-2(t-5)}  - e^{-5(t - 5)} ]

⇒y(t) =  3u₂(t) [ e^{-2t+4}  - e^{-5t + 10)} ] - 4u₅(t) [ e^{-2t+10)}  - e^{-5t + 25)} ]

It is the required solution.

3 0
3 years ago
Simplify 6(x + 4) 10x 6x + 4 6x + 24
Maurinko [17]

Answer:

=360x4+1440x3+24x+24

Step-by-step explanation:

6(x+4)(10)xx(6)x+(4)(6)x+24

=360x4+1440x3+(4)(6)x+24

=360x4+1440x3+24x+24

Hope it helps!!

6 0
3 years ago
Read 2 more answers
Which fraction is simplified fraction of 28/56
stepan [7]

Answer:

1/2

Step-by-step explanation:

8 0
4 years ago
Read 2 more answers
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